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Derivative proof

  1. Nov 4, 2008 #1
    1. The problem statement, all variables and given/known data
    is a) fg''-f''g=(fg'-f'g)'
    b) fg''+f''g=(fg)''


    2. Relevant equations
    product formula? f'g+fg'=fg'
    quotient rule- (f'g-fg')/g^2=(f/g)'



    3. The attempt at a solution
    All i did was plug in numbers, such that f=x^3 and g=x^2 and then solved it that way. For doing this, I got that a is true and that b is false, but it isn't really a proof if you use numbers, hell it isn't a proof at all. :smile: Any ideas on how to make it so that it either IS or ISNT true.
     
  2. jcsd
  3. Nov 4, 2008 #2

    Mark44

    Staff: Mentor

    Instead of "plugging in numbers" (you're actually plugging in functions, not numbers), just use the product rule.

    Problem a (reading from right to left) says that the derivative of fg' - f'g is what is shown on the left. What is the derivative of fg' - f'g.

    Problem b (again reading from right to left) says that the 2nd derivative of fg is shown on the left. Take the derivative of fg and see what you get. Take the derivative of that and see what you get.
     
  4. Nov 4, 2008 #3
    Mark,
    What i dont understand is how can i just take the derivative of f and g?
    We know all the rules, but i dont get what you mean by take the derivative of it.
     
  5. Nov 4, 2008 #4
    Sorry mark,
    I completely understand now, makes no sense why i didnt do it before. I have been making stupid mistaes like this for the past two days, I have a huge problem set due for math physics and a big research paper due, so im kinda just burnt out i guess. Thanks tho for the nudge mark.


    just to check, i got the first one being equal, second one not being equal?
     
  6. Nov 5, 2008 #5

    Mark44

    Staff: Mentor

    I didn't work the first one, but it's pretty easy now that you know how. I worked the second one and agree that the two expressions aren't equal.
     
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